2次方程式 $2x^2 + 3x - 1 = 0$ の2つの解を $\alpha$ と $\beta$ とするとき、以下の値をそれぞれ求めよ。 (1) $\alpha^2 \beta + \alpha \beta^2$ (2) $\frac{1}{\alpha+1} + \frac{1}{\beta+1}$ (3) $(\alpha - 1)(\beta - 1)$

1. 問題の内容

2次方程式 $2x^2 + 3x - 1 = 0$ の2つの解を $\alpha$ と $\beta$ とするとき、以下の値をそれぞれ求めよ。

(1) $\alpha^2 \beta + \alpha \beta^2$

(2) $\frac{1}{\alpha+1} + \frac{1}{\beta+1}$

(3) $(\alpha - 1)(\beta - 1)$

2. 解き方の手順

解と係数の関係より、

$\alpha + \beta = -\frac{3}{2}$

$\alpha \beta = -\frac{1}{2}$

(1) $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta)$

$\alpha^2 \beta + \alpha \beta^2 = (-\frac{1}{2})(-\frac{3}{2}) = \frac{3}{4}$

(2) $\frac{1}{\alpha+1} + \frac{1}{\beta+1} = \frac{(\beta+1) + (\alpha+1)}{(\alpha+1)(\beta+1)} = \frac{\alpha + \beta + 2}{\alpha\beta + \alpha + \beta + 1}$

$\frac{1}{\alpha+1} + \frac{1}{\beta+1} = \frac{-\frac{3}{2} + 2}{-\frac{1}{2} - \frac{3}{2} + 1} = \frac{\frac{1}{2}}{-1+1} = \frac{\frac{1}{2}}{-1}$

分母が0になるので、計算を見直す。

$\frac{1}{\alpha+1} + \frac{1}{\beta+1} = \frac{\beta+1 + \alpha+1}{(\alpha+1)(\beta+1)} = \frac{\alpha + \beta + 2}{\alpha\beta + \alpha + \beta + 1}$

$\frac{1}{\alpha+1} + \frac{1}{\beta+1} = \frac{-\frac{3}{2}+2}{-\frac{1}{2} - \frac{3}{2}+1} = \frac{\frac{1}{2}}{-\frac{4}{2}+1+\frac{2}{2}} = \frac{\frac{1}{2}}{-2+1} = \frac{\frac{1}{2}}{-1} = -\frac{1}{2}$

(3) $(\alpha - 1)(\beta - 1) = \alpha \beta - (\alpha + \beta) + 1$

$(\alpha - 1)(\beta - 1) = -\frac{1}{2} - (-\frac{3}{2}) + 1 = -\frac{1}{2} + \frac{3}{2} + 1 = \frac{2}{2} + 1 = 1+1 = 2$

3. 最終的な答え

(1) $\frac{3}{4}$

(2) $-\frac{1}{2}$

(3) $2$